The 8903 will display its input from the amp in volts RMS, watts, or relative dB - or calculate distortion, (S+N)/N, etc. Although the 8903's input is always high impedance and thus doesn't load down the amp, you can set the 8903 according to impedance of your dummy load so that it calculates the power correctly. Its input range is 300VRMS, which is more that 10,000 watts in an 8 ohm system.
The 8903B's input range being up to 300Vrms makes sense to me now. This means the dummy load is easier to make. The issue with the 8903B is it's lowest THD measurement seems to be 0.01% (per
this doc). Although there are reports of getting down to 0.005%.
You may want to download Steven Tate's write-up on the QuantAsylum QA401:
Steven Tate's QA401 Write-up
Here's the link to the Write-up:
QA401 Write-up
Nice write-up to get someone up to speed on setting it up and using it.
Paul
Paul, thanks for posting this! I hadn't seen this document before. I've read it and I'm trying to work out the best config for a dummy load for the QA403. Below is my line of thinking.
Formulas:
Vrms = 10^(dBV / 20)
Vrms = Square Root (Power Watts * System Z)
dBV = 20*Log10(Vrms/1V)
W = V^2/R
Voltage Divider: Vout = (V1*R2)/(R1+R2)
QA 403 has in input level max of 32dBV (40Vrms, 200W @ 8 Ohm). The better idea is to run it at 80% of that rating which is 25.6dBV (19.05Vrms, 45W @ 8 Ohm)
I have a 120W (into 8 ohm) power amp (Marantz 240) that I'd like to measure. So, let's do some math:
Vrms = Square Root (Power Watts * System Z)
Vrms = sqrt (120 * 8)
Vrms = 30.98
dBV = 20*Log10(30.98/1V)
dBV = 29.82
29.82 dBV is above the 80% threshold for the QA 403 (25.6dBV). So according to Steven's document, adding those two 5.6k ohm resistors in series should give us a voltage divider which removes 6dBV (half the voltage). So that would get us to ~23dBV. This method would meet my current need. What would someone do if they had a 200W amp (into 8 ohms)? Is this where the video from VAR comes in?
Using the above formula, 200W into 8 Ohms is 40Vrms which is ~32dBV. So again, too high for the 80% threshold of the QA403. If we use the same 5.6k Ohm resistors, we bring the voltage down to 20Vrms (assuming this formula works for RMS Volts). Which is slightly above the 80% limit for the QA403:
Vout = (V1*R2)/(R1+R2)
Vout = (40*5600)/(5600+5600)
Vout = 20
20Vrms = 26dBV
This is probably acceptable.
VAR is "tapping" into his resistor series to lower the dBV for his QA402 after 7 Ohms.
Vout = (40*1)/(7+1)
Vout = 5Vrms
5Vrms = 13.979dBV
32dBV - 13.9dBV = 18.1dBV (as shown in his video)
An 800W amp into 8 Ohms = 38dBV. So tapping into the -18dBV position on the dummy load would yield 20dBV, which is within the acceptable range.
The max wattage that the QA403 could handle using the VAR attenuation method would be:
25.6dBV + 18dBV = 43.6dbV
Vrms = 151.4
W (into 8 Ohms) = 2863.6
That's several high wattage resistors!
The downside of VAR's approach is that he's measuring distortion at a suboptimal location. I suppose the 8903 is doing the same thing, albeit, on it's internal attenuators. I guess there's really no way of getting around measuring high voltage power amps with "taps" (voltage divider) on the dummy load. Am I missing anything in these calcs? Is there a different way to lower the input voltage for the QA 403? It seems like the 403 is the better option for more accurate distortion measurements. I just like the thought of having a 8903B on the bench! Thanks to all who responded. Sorry for the late reply. Weekends are family time!
Jim